Question
An observer, $1.7m$ tall, is $20\sqrt{3}\text{m}$ away from a tower. The angle of elevation from the eye of an observer to the top of tower is $30^\circ$. Find the height of the tower.
✓
Answer
$\text{CB}=\text{DE}=20\sqrt{3}\text{m}$
In $\triangle\text{ABC,}$
$\frac{\text{AB}}{\text{BC}}=\tan30^\circ$
$\frac{\text{H}-1.7}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$
$(\text{H}-1.7)\sqrt{3}=20\sqrt{3}$
$\text{H}-1.7=20$
$\text{H}=20+1.7=21.7\text{m}$
In $\triangle\text{ABC,}$
$\frac{\text{AB}}{\text{BC}}=\tan30^\circ$
$\frac{\text{H}-1.7}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$
$(\text{H}-1.7)\sqrt{3}=20\sqrt{3}$
$\text{H}-1.7=20$
$\text{H}=20+1.7=21.7\text{m}$
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